TITLE: How can we prove that $[ {c^n - (a^n+b^n)} + d^n ] / (ab-cd)$ will yield integers always?
QUESTION [6 upvotes]: Let us say $a<b<c$ and $(a+b)=(c+d)$ as well as $n\geq2$. Then it is found for all values of $n$ that 
  $$
  \frac{{c^n-(a^n+b^n)}+d^n}{ab-cd} \in \mathbb{Z}
$$

Can you explain the reason for this?

REPLY [3 votes]: First we give a simple inductive proof then we give a conceptual interpretation of the induction as a special case of the uniqueness theorem for recurrences (difference equations).
Hint $\ $ The inductive step of a simple inductive proof follows by subtracting these
$$\begin{align}  
c^{n+2}+d^{n+2} &= (c+d)(c^{n+1}+d^{n+1}) - cd(c^n+d^n)\\
a^{n+2}+b^{n+2} &= (a+b)(a^{n+1}+b^{n+1}) - ab(a^n+b^n)\end{align}\qquad$$
Indeed $\,{\rm mod}\ ab\!-\!cd\!:\ $ by induction $\,c^k\!+d^k\equiv a^k\!+b^k\,$ for $\,k = n,\,n\!+\!1\,$ so subtracting
$$c^{n+2}\!+\!d^{n+2}\!-\!a^{n+2}\!-\!b^{n+2}\equiv \underbrace{(c\!+\!d\!-\!a\!-\!b)}_{\large =\, 0}(a^{n+1}\!+b^{n+1}) + \underbrace{(ab\!-\!cd)}_{\large \equiv\, 0}(a^n\!+b^n)\,\equiv\, 0$$
Remark $\ $ More generally this yields $\ \gcd(c\!+\!d\!-\!a\!-\!b,ab\!-\!cd)\mid c^n\!+d^n\!-a^n\!-b^n$
Conceptually $\ f_n = a^n\!+b^n$ and $\,{\bar f}_{\!n} =  c^n+d^n$ are both solutions of the same recurrence $\,  f_{n+2} = A f_{n+1} - B f_n\,$ for $\, a+b\equiv A\equiv c+d,\ $ $\,ab \equiv B\equiv cd,\,$ with same initial conditions $\,f_0 \equiv 2,\ f_1 \equiv a+b\equiv c+d,\,$ thus $\,f_n\equiv {\bar f}_{\!n}\,$ for all $\,n\,$ by the same simple induction as above. This is just  the uniqueness theorem for recurrences.